I was thinking over the following statement:
Given a Noetherian ring $R$ and a bounded cochain complex $ C^{\bullet}$ with finitely generated cohomology, that is, $H^{i}(C^{\bullet})$ is finitely generated for all $i$.
Then there exists a bounded cochain complex $B^{
\bullet} $of finitely generated $R$-modules is quasi-isomorphic to $C^{\bullet}$.
This is a part (essential surjectivity) of the statement that the natural functor
$$D^b(\mathrm{mod-}R)\longrightarrow D^b_{\mathrm{mod-}R}(\mathrm{Mod-}R)$$
is an equivalence, where $\mathrm{mod-}R$ is the category of finitely generated $R$-modules and $\mathrm{Mod-}R$ is the category of $R$-modules. For more details, see stacks project 06UP (Lemma 13.17.4.).
There are some proof on stacks project and Sheaves on Manifolds. But neither of them make sense to me…
For the one on Sheaves on Manifolds, we need the condition that
for any monomorphism $M\to N$ with $M$ is finitely generated, we have a finitely generated module $N'$ and a morphism $N\to N'$ such that the composition $M\to N\to N'$ is a monomorphism.
However, I fail to show the condition holds.
For the one on stacks project, given a bounded complex
$$X^\bullet: 0\to X^0\to X^1\to \cdots \to X^{n-1}\to X^n \to 0,$$
taking $B^\bullet=0$ in the proof, we can get $C^\bullet$ such that $H^{n}(C^\bullet)\to H^n(X^\bullet)$ is surjective. But I fail to see how the complex $E^\bullet$ can make the map to be injective.
If there any good reference for the above statement ? Is the Noetherian condition necessary?
Any answer and hint are welcome!
Update: I tried to use the dual version of the proof on the Sheaves on Manifolds, and it worked.
Best Answer
It looks from your edit as though you now know how to use the results from Sheaves on Manifolds to get the result you want.
But if you're specifically interested in the fact about bounded complexes of finitely generated modules for a Noetherian ring, then another method is simply to take a projective resolution of your complex and truncate it in sufficiently small degree.
It is not hard to check that if the cohomology of your complex is finitely generated in each degree and bounded, then you can construct a projective resolution that is finitely generated in each degree. Or alternatively, to check that the subcategory of $D^b(\text{Mod-}R)$ consisting of complexes with projective resolutions finitely generated in each degree is a triangulated subcategory that contains all finitely generated modules, and so contains $D^b_{\text{mod-}R}(\text{Mod-}R)$.